Problem: Suppose we have a plane $P$ defined by the transformation $T$ for $4 < u < 5$ and $0 < v < 2$. $T(u, v) = (3, 2u + 3v, -u + 2v)$ What is the surface area of $P$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $14$ (Choice B) B $2\sqrt{7}$ (Choice C) C $7$ (Choice D) D $2\sqrt{14}$
Explanation: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_0^2 \int_4^5 |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = 7$ [Calculation] The final step is to evaluate the double integral. $\begin{aligned} A &= \int_0^2 \int_4^5 |T_u \times T_v| \, du \, dv \\ \\ &= \int_0^2 \int_4^5 7 \, du \, dv \\ \\ &= \int_0^2 7 \, dv \\ \\ &= 14 \end{aligned}$ In conclusion, the surface area of $P$ is $14$.